   1. Chapter 12 Class 11 Introduction to Three Dimensional Geometry
2. Serial order wise
3. Ex 12.3

Transcript

Ex 12.3, 2 Given that P (3, 2, –4), Q (5, 4, –6) and R (9, 8, –10) are collinear. Find the ratio in which Q divides PR. Given that Point P (3, 2, –4), Q (5, 4, –6) & R (9, 8, –10) are collinear Q must divide line segment PR in some ratio externally & internally . We know that Co-ordinate of point P(x, y ,z) that divides line segment joining (x1, y1, z1) & (x2, y2, z2) in ration m : n is (x, y ,z) = ﷐﷐mx2 + nx1﷮m + n﷯,﷐my2+ my1﷮m + n﷯﷯ Let point Q (5, 4, –6) divide line segment P (3, 2, – 4), R (9, 8, – 10) in the ratio k : 1 Here, x1 = 3, y1 = 2, z1 = – 4 x2 = 9, y2 = 8, z2 = – 10 & m = k , n = 1 Putting values Q (5, 4, – 6) = ﷐﷐k﷐9﷯+3﷮k+1﷯,﷐k﷐8﷯+2﷮k+1﷯,﷐k﷐−10﷯−4﷮k+1﷯﷯ (5, 4, – 6) = ﷐﷐9𝑘 + 3﷮k + 1﷯,﷐8𝑘 + 2﷮k + 1﷯,﷐− 10𝑘 −4﷮k+1﷯﷯ Comparing x – coordinate of Q 5 = ﷐9k + 3﷮k + 1﷯ 5 (k + 1) = 9k + 3 5k + 5 = 9k + 3 5k – 9k = 3 – 5 – 4k = – 2 k = ﷐−2﷮−4﷯ k = ﷐1﷮2﷯ So, k : 1 = 1 : 2 Thus, point Q divides PR in the Ratio 1 : 2

Ex 12.3 